Thực hiện phép tính
( x - 1).( x +1). ( x + 2)
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1: ĐKXĐ: \(x\notin\left\{0;3\right\}\)
\(\dfrac{x+3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\)
\(=\dfrac{x+3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x\left(x-3\right)}\)
\(=\dfrac{\left(x+3\right)\left(x-3\right)-x^2+9}{x\left(x-3\right)}\)
\(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}\)
=0
2: ĐKXĐ: \(x\notin\left\{0;1\right\}\)
\(\dfrac{3}{x}-\dfrac{5}{x-1}+\dfrac{3x+2}{x^2-x}\)
\(=\dfrac{3}{x}-\dfrac{5}{x-1}+\dfrac{3x+2}{x\left(x-1\right)}\)
\(=\dfrac{3x-3-5x+3x+2}{x\left(x-1\right)}\)
\(=\dfrac{x-1}{x\left(x-1\right)}=\dfrac{1}{x}\)
1 ( 131,4 - 80,8 ) : 2,3 + 21,84 x 2 = 50,6 : 2,3 + 43,68 = 22 + 43,68= 65,68
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot....\cdot\left(1-\frac{1}{2003}\right)\cdot\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot....\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot4\cdot....\cdot2002\cdot2003}{2\cdot3\cdot4\cdot5\cdot....\cdot2003\cdot2004}\)
\(=\frac{1}{2004}\)
\(\dfrac{5x+5y}{3x-3y}:\dfrac{5x}{x^2-y^2}.\)
\(=\dfrac{5\left(x+y\right)}{3\left(x-y\right)}.\dfrac{\left(x-y\right)\left(x+y\right)}{5x}.\)
\(=\dfrac{x+y}{3}.\dfrac{x+y}{x}.\)
\(=\dfrac{\left(x+y\right)^2}{3x}.\)
Bài làm
1) 29( 85 - 47 ) + 85( 47 - 29 )
= 29 . 85 - 29 . 47 + 85 . 47 - 29 . 85
= 29 . 85 . ( -29 . 47 + 85 . 47 )
= 29 . 85 . [ 47( -29 + 82 ) ]
= 29 . 85 . ( 47 . 53 )
= 6140315
2) 3 - | x - 7 | = -12
=> | x - 7 | = 3 - ( -12 )
=> | x - 7 | = 15
=> \(\orbr{\begin{cases}x-7=15\\x-7=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=22\\x=-8\end{cases}}}\)
Vậy x = { 22; -8 }
Ta có:
\(\left(x-1\right).\left(x+1\right).\left(x+2\right)\)
\(=\left(x^2+x-x-1\right).\left(x+2\right)\)
\(=\left(x^2-1\right).\left(x+2\right)\)
\(=x^3+2x^2-x-2\)